比较两个数组并过滤它们javascript

Question

Basically i have two arrays which i am using it for the configuration of buttons.

First array which defines how many buttons should be present and in order.

buttonGroups: [ 0, 2 ]

Another array of objects which says about the actual buttons.

    buttons = [
    {
        buttonLabel: 'label1',
        cond1: true,
        cond2: false
    },
    {
        buttonLabel: 'label2',
        cond1: true,
        cond2: false
    },
    {
        buttonLabel: 'label3',
        cond1: false,
        cond2: true
    }
];

The buttonGroups is the configuration array. If it has only [0, 1] then first two buttons will exist. If buttonGroups has only [0, 3] we should have first and third button exists in buttons array.

This is what i have tried

buttonGroups.map((payload1, index1) => {
    buttons .map((payload2, index2) => {
        if(index1 === index2){
            //Display Here only the matched index from ButtonGroups
            console.log(payload2)
        }
    })
})

This is giving the first index button array. How to get the matched array buttons?

Answer 1

Here you go with a solution

 var buttonGroups = [ 0, 2 ]; var buttons = [ { buttonLabel: 'label1', cond1: true, cond2: false }, { buttonLabel: 'label2', cond1: true, cond2: false }, { buttonLabel: 'label3', cond1: false, cond2: true } ]; var filteredButtons = buttonGroups.map(item => { return buttons[item]; }); console.log(filteredButtons);

filteredButtons will return the filtered buttons which you can render.

Answer 2

您可以遍历buttonGroups并获得结果：

buttonGroups.map(button => {return buttons[button]})

Answer 3

使用filter方法。

const filterBtn = buttons.filter((btn,index) => buttonGroups.includes(index));